How do you solve #(8x+3)(x+3)=0#?

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Answer 1 · 2017-09-15T21:05:39

See a solution process below:

Equate each term in parenthesis on the left side of the equation to #0# and solve for #x#:

Solution 1:

#8x + 3 = 0#

#8x + 3 - color(red)(3) = 0 - color(red)(3)#

#8x + 0 = -3#

#8x = -3#

#(8x)/color(red)(8) = -3/color(red)(8)#

#(color(red)(cancel(color(black)(8)))x)/cancel(color(red)(8)) = -3/8#

#x = -3/8#

Solution 2:

#x + 3 = 0#

#x + 3 - color(red)(3) = 0 - color(red)(3)#

#x + 0 = -3#

#x = -3#

The Solutions Are: #x = -3/8# and #x = -3#